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  • Can someone help me with this, A stream of electrons from a heated filament was passed between two charged plates kept at a potential difference V esu. If e and m are charge and mass of an electron, respectively, then the value of h/λ (where

A stream of electrons from a heated filament was passed between two charged plates kept at a potential difference V esu. If e and m are charge and mass of an electron, respectively, then the value of h/λ (where λ is wavelength associated with electron wave) is given by

  • Option 1)

    meV

  • Option 2)

    2meV

  • Option 3)

    \sqrt{meV}

  • Option 4)

    \sqrt{2meV}

 

Answers (2)

As we have learned

De-broglie wavelength -

\lambda = \frac{h}{mv}= \frac{h}{p}

- wherein

where m is the mass of the particle

v its velocity 

p its momentum

 

 De Brogile wavelength \lambda = \frac{h}{p} \\

p= \frac{h}{\lambda}.......(1)

Kinetic energy of an electron in eV 

K.E = \frac{p^2}{2m}

\therefore eV = \frac{p^2}{2m}

Or p = \sqrt{2meV}......(2)

from equtaion (1)  and (2) 

\frac{h}{\lambda }= \sqrt{2meV}

 

 

 

 

 


Option 1)

meV

Option 2)

2meV

Option 3)

\sqrt{meV}

Option 4)

\sqrt{2meV}

Posted by

Vakul

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